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Calculate the approximate total Ca conc. in the river water contributed by the Ca(OH)2 (provide all your assumptions) (Given Ca(OH)2 (s) Ksp = 10 -5.43 and Ca(OH)2 (s) = CaOH+ + OH-, Keq = 10-4.03.

 
Apr 11th, 2015

Ksp = [Ca2+] * [OH-] 

Ca(OH)2 ↔ Ca2+ + 2OH- 
Let X = [Ca2+] 

Ksp= [X]*[2X]^2

10^-5.43= 4X^3

X=0.0098

so [Ca2+]= 9.8*10^-3M................................best me

Apr 11th, 2015

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Apr 11th, 2015
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